What do you mean by "the" solution? Such an equation will have infinitely many solutions. In fact, I would expect to have three arbitrary functions, determined by two "boundary conditions" and one "initial condition".
As Subhotosh Kahn suggests, you can try "separation of variables": let h(r, t)= U(r)V(t), a product of a function of r only and a function of t only. Then the equation becomes \(\displaystyle U\frac{dV}{dt}= \frac{S}{T}V\frac{d^2U}{dr^2}\). Divide both sides by UV to get \(\displaystyle \frac{1}{V}\frac{dV}{dt}= \frac{S}{T}\frac{1}{U}\frac{d^2U}{dr^2}\).
Now, the left side is a function of t only while the right side is a function of r only. The only way they can be equal for all t and r is if each is a constant: \(\displaystyle \frac{1}{V}\frac{dV}{dt}= \frac{S}{T}\frac{1}{U}\frac{d^2U}{dr^2}= \alpha\) for \(\displaystyle \alpha\) any constant. That we can write as two separate equations, \(\displaystyle \frac{dV}{dt}= \alpha V\) and \(\displaystyle \frac{d^2U}{dr^2}= \frac{S}{T}\alpha U\).
Now, how we would solve those equations, indeed, what \(\displaystyle \alpha\) could be, depend strongly upon the boundary conditions on U and the initial condition on V.